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Chemical Thermodynamics Formulas For JEE 2026
Chemical Thermodynamics is an important chapter in JEE Chemistry because it explains how energy changes take place during chemical reactions. In this topic, students learn about key concepts such as the laws of thermodynamics, enthalpy, entropy, and Gibbs free energy. These ideas help in understanding how heat and energy are involved in different chemical processes and how they determine whether a reaction can occur.
Many JEE questions are based on applying thermodynamic concepts to solve problems related to heat changes, spontaneity, and chemical equilibrium. When students clearly understand these formulas and principles, solving numerical problems becomes much easier. For quick revision during preparation, students can also refer to a well-structured JEE Mains Chemistry Formula PDF, which helps them review important formulas and key concepts efficiently.
First Law of Thermodynamics
Types of Systems
| Type | Energy Exchange? | Matter Exchange? |
|---|---|---|
| Open System | Yes | Yes |
| Closed System | Yes | No |
| Isolated System | No | No |
First Law of Thermodynamics
$$$\Delta U = q + w$$$
where $$\Delta U$$ = change in internal energy, $$q$$ = heat exchanged, $$w$$ = work done.
For expansion/compression work against external pressure:
$$$w = -P_{\text{ext}} \Delta V = -P_{\text{ext}}(V_2 - V_1)$$$
So: $$\Delta U = q - P_{\text{ext}} \Delta V$$
Sign Convention
- Heat ($$q$$): Positive if absorbed by the system (endothermic); negative if released (exothermic)
- Work ($$w$$): Positive if done on the system; negative if done by the system
Worked Example: First Law
A gas absorbs 500 J of heat and expands, doing 200 J of work on the surroundings. Find $$\Delta U$$.
$$q = +500$$ J, $$w = -200$$ J (work done by system)
$$\Delta U = q + w = 500 + (-200) = \textbf{+300 J}$$
Work in Different Processes
Types of Thermodynamic Processes
- Isothermal (constant $$T$$): $$\Delta T = 0$$
- Isobaric (constant $$P$$): $$\Delta P = 0$$
- Isochoric (constant $$V$$): $$\Delta V = 0$$, so $$w = 0$$ and $$\Delta U = q_V$$
- Adiabatic (no heat exchange): $$q = 0$$, so $$\Delta U = w$$
Work Done by an Ideal Gas
Irreversible (against constant external pressure):
$$$w = -P_{\text{ext}}(V_2 - V_1)$$$
Reversible isothermal expansion/compression:
$$$w = -nRT \ln\frac{V_2}{V_1} = -2.303 \, nRT \log\frac{V_2}{V_1}$$$
Worked Example: Reversible Isothermal Work
2 mol of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 20 L. Calculate the work done.
$$w = -nRT \ln\dfrac{V_2}{V_1} = -(2)(8.314)(300) \ln 2 = -4988.4 \times 0.693 = \textbf{-3457 J} \approx -3.46$$ kJ
Enthalpy
Enthalpy Change
$$$H = U + PV$$$
At constant pressure:
$$$\Delta H = \Delta U + P\Delta V \qquad \text{and} \qquad \Delta H = q_P$$$
For reactions involving ideal gases:
$$$\Delta H = \Delta U + \Delta n_g RT$$$
where $$\Delta n_g$$ = (moles of gaseous products) − (moles of gaseous reactants).
Worked Example: $$\Delta H$$ and $$\Delta U$$
For $$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$$, if $$\Delta H = -92.4$$ kJ at 298 K, find $$\Delta U$$.
$$\Delta n_g = 2 - (1 + 3) = -2$$
$$\Delta U = \Delta H - \Delta n_g RT = -92.4 - (-2)(8.314 \times 10^{-3})(298) = -92.4 + 4.955 = \textbf{-87.4 kJ}$$
Exothermic vs Endothermic
| Property | Exothermic | Endothermic |
|---|---|---|
| Heat flow | Released to surroundings | Absorbed from surroundings |
| $$\Delta H$$ | Negative ($$< 0$$) | Positive ($$> 0$$) |
| Example | Combustion of fuels | Melting of ice |
Standard Enthalpy Changes
Using Formation Enthalpies
$$$\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})$$$
$$\Delta_f H°$$ of elements in their standard state = 0 (by definition).
Worked Example: Combustion of Methane
Calculate $$\Delta_r H°$$ for: $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$
$$\Delta_r H° = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] = -965.1 + 74.8 = \textbf{-890.3 kJ/mol}$$
Enthalpy from Bond Energies
$$$\Delta_r H° = \sum (\text{Bond energies of bonds broken}) - \sum (\text{Bond energies of bonds formed})$$$
Breaking bonds = positive (endothermic). Forming bonds = negative (exothermic).
Worked Example: Bond Energies
Estimate $$\Delta_r H°$$ for: $$\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$$
Bond energies: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, H–Cl = 431 kJ/mol
Bonds broken: 436 + 242 = 678 kJ. Bonds formed: 2 × 431 = 862 kJ.
$$\Delta_r H° = 678 - 862 = \textbf{-184 kJ/mol}$$ (exothermic)
Hess's Law and Kirchhoff's Equation
Hess's Law
$$$\Delta H_{\text{reaction}} = \Delta H_1 + \Delta H_2 + \cdots$$$
Rules:
- If a reaction is reversed, the sign of $$\Delta H$$ changes.
- If a reaction is multiplied by a factor $$n$$, $$\Delta H$$ is also multiplied by $$n$$.
Kirchhoff's Equation
$$$\Delta H_2 = \Delta H_1 + \Delta C_P (T_2 - T_1)$$$
where $$\Delta C_P = \sum C_P(\text{products}) - \sum C_P(\text{reactants})$$
Entropy and the Second Law
Entropy Change
$$$\Delta S = \frac{q_{\text{rev}}}{T}$$$
For a reaction: $$\Delta_r S° = \sum S°(\text{products}) - \sum S°(\text{reactants})$$
Second Law of Thermodynamics
$$$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$$$
- $$\Delta S_{\text{universe}} > 0$$: Process is spontaneous
- $$\Delta S_{\text{universe}} = 0$$: Process is at equilibrium
- $$\Delta S_{\text{universe}} < 0$$: Process is non-spontaneous
Predicting Entropy Changes
Entropy increases ($$\Delta S > 0$$) when:
- Solid → liquid → gas (phase change increasing disorder)
- Number of gas moles increases ($$\Delta n_g > 0$$)
- Temperature increases
- A solute dissolves in a solvent
Gibbs Free Energy
Gibbs–Helmholtz Equation
$$$\Delta G = \Delta H - T\Delta S$$$
- $$\Delta G < 0$$: Reaction is spontaneous
- $$\Delta G = 0$$: Reaction is at equilibrium
- $$\Delta G > 0$$: Reaction is non-spontaneous
Effect of $$\Delta H$$ and $$\Delta S$$ on Spontaneity
| $$\Delta H$$ | $$\Delta S$$ | $$\Delta G$$ | Spontaneity |
|---|---|---|---|
| $$-$$ | $$+$$ | Always negative | Spontaneous at all $$T$$ |
| $$+$$ | $$-$$ | Always positive | Non-spontaneous at all $$T$$ |
| $$-$$ | $$-$$ | Negative at low $$T$$ | Spontaneous at low $$T$$ |
| $$+$$ | $$+$$ | Negative at high $$T$$ | Spontaneous at high $$T$$ |
Worked Example: Spontaneity
For a reaction, $$\Delta H = -10.5$$ kJ/mol and $$\Delta S = +31.4$$ J/(mol·K). Is it spontaneous at 298 K?
$$\Delta G = -10500 - (298)(31.4) = -10500 - 9357.2 = \textbf{-19857 J/mol} = -19.86$$ kJ/mol
Since $$\Delta G < 0$$, the reaction is spontaneous at 298 K.
Worked Example: Crossover Temperature
At what temperature does the decomposition of CaCO$$_3$$ become spontaneous?
Given: $$\Delta H = +178.3$$ kJ/mol, $$\Delta S = +160.4$$ J/(mol·K)
At threshold: $$\Delta G = 0$$, so $$T = \dfrac{\Delta H}{\Delta S} = \dfrac{178300}{160.4} = \textbf{1112 K} \approx 839°$$C
Tip: When $$\Delta H$$ and $$\Delta S$$ have the same sign, there is a crossover temperature $$T = \Delta H / \Delta S$$ where $$\Delta G = 0$$. JEE often asks you to find this temperature.
Gibbs Free Energy and Equilibrium
$$\Delta G$$ and Equilibrium Constant
$$$\Delta G = \Delta G° + RT \ln Q$$$
At equilibrium ($$\Delta G = 0$$, $$Q = K$$):
$$$\Delta G° = -RT \ln K = -2.303 \, RT \log K$$$
- $$\Delta G° < 0$$: $$K > 1$$ (products favoured)
- $$\Delta G° = 0$$: $$K = 1$$
- $$\Delta G° > 0$$: $$K < 1$$ (reactants favoured)
Worked Example: Equilibrium Constant from $$\Delta G°$$
Calculate $$K$$ at 298 K for a reaction with $$\Delta G° = -13.6$$ kJ/mol.
$$-13600 = -2.303 \times 8.314 \times 298 \times \log K$$
$$\log K = \dfrac{13600}{5705.8} = 2.383$$
$$K = 10^{2.383} \approx \textbf{241.5}$$
Summary of Key Formulas
Quick Reference
- First Law: $$\Delta U = q + w$$
- Enthalpy: $$\Delta H = \Delta U + \Delta n_g RT$$
- Reversible isothermal work: $$w = -nRT \ln(V_2/V_1)$$
- Hess's Law: $$\Delta H_{\text{rxn}} = \sum \Delta H_{\text{steps}}$$
- Kirchhoff: $$\Delta H_2 = \Delta H_1 + \Delta C_P(T_2 - T_1)$$
- Entropy: $$\Delta S = q_{\text{rev}} / T$$
- Gibbs: $$\Delta G = \Delta H - T\Delta S$$
- Equilibrium: $$\Delta G° = -RT \ln K$$
Important Note
The three most important equations for JEE from this chapter are: (1) $$\Delta U = q + w$$ (First Law), (2) $$\Delta H = \Delta U + \Delta n_g RT$$ (relating enthalpy and internal energy), and (3) $$\Delta G = \Delta H - T\Delta S$$ (Gibbs equation for spontaneity). Master these three and you can solve most thermodynamics problems.
Chemical Thermodynamics Formulas For JEE 2026: Conclusion
Chemical thermodynamics is an important topic that helps students understand how energy changes occur during chemical reactions. Concepts such as internal energy, enthalpy change, entropy, and Gibbs free energy explain whether a reaction will release energy, absorb energy, or occur spontaneously. When these principles are clearly understood, solving thermodynamics problems in competitive exams becomes much easier.
In addition, ideas like Hess’s law, Kirchhoff’s equation, and equilibrium relationships help students analyze reactions from an energy perspective. Regular practice and quick revision of key formulas can improve problem-solving speed and accuracy, which is very useful while preparing for competitive exams like JEE.
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